Question - Determining the Value of an Angle in a Quadrilateral
Solution:
Given:- Quadrilateral $$ ABCD $$- $$ AB = BC = CD = 4 $$- $$ \angle C = 32^\circ $$- $$ \angle A = 115^\circ $$- $$ BD $$ is a diagonalTo find:- $$ \angle x $$ in $$ \triangle BCD $$Step 1: Identify that $$ ABCD $$ is a kite since two pairs of adjacent sides are equal.Step 2: Use the fact that in a kite, one diagonal is the perpendicular bisector of the other.Step 3: This makes $$ \triangle BCD $$ and $$ \triangle ABD $$ right triangles. Therefore, $$ \angle BDC = \angle ABD = 90^\circ $$.Step 4: Calculate the remaining angle in $$ \triangle BCD $$ (angle $$ x $$) using the fact that the sum of angles in a triangle is $$ 180^\circ $$.\[ x + 32^\circ + 90^\circ = 180^\circ \]\[ x + 122^\circ = 180^\circ \]\[ x = 180^\circ - 122^\circ \]\[ x = 58^\circ \]
Tutor Solution:
Given the quadrilateral $$ABCD$$, it is necessary to solve for the value of $$x$$. In a quadrilateral, the sum of the internal angles equals $$360^\circ$$. We can use this information along with the angles provided to find $$x$$.First, let us summarize the angles we have:- $$ \angle A = 115^\circ $$- $$ \angle B = 115^\circ $$ (since $$ AB = BD $$, triangle $$ABD$$ is isosceles and angles opposite to equal sides are also equal)- $$ \angle C = 32^\circ $$- $$ \angle D = x^\circ $$Adding together the angles we know and setting them equal to $$360^\circ$$, we get the equation:\[115^\circ + 115^\circ + 32^\circ + x^\circ = 360^\circ\]Now, we simplify:\[262^\circ + x^\circ = 360^\circ\]To isolate $$x$$, we subtract $$262^\circ$$ from both sides:\[x^\circ = 360^\circ - 262^\circ\]Therefore, the value of $$x$$ is:\[x = 98^\circ\]Hence, angle $$D$$ measures $$98^\circ$$.
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